Question

Industrial ethanol (CH3CH2OH) is produced by a catalytic reaction of ethylene (CH2═CH2) with water at high pressures and temperatures. Calculate ΔH o rxn for this gas-phase hydration of ethylene to ethanol, using bond energies and then using enthalpies of formation.

Answer 1

Answer:

using bond energies = - 37 kJ

using enthalpies of formation = - 45.7 kJ

Explanation:

From the reaction;

CH2═CH2(g) + H2O(g) -------> CH3CH2OH(g)

From the above reaction; there are 4 (C-H) bonds , 1 ( C=C) and 2 (H-O) of ethene which forms 5(C-H) bonds, 1 (C-C) and 1 (C-O) and 1(H-O) bonds.

Using Bond Energies; the heat of the reaction can be written as:

$\delta H^0_ {rxn}=$ ∑ energy of old bond breaking + ∑ energies of the new bond formation.

$\delta H^0_ {rxn}=$ ∑ $[(4 * BE_{C-H}) + BE_{C-C}) + + BE_{O-H})]$+ ∑ $[(5 * BE_{C-H}) + BE_{C-C}) + BE_{O-H}+ BE_{C-O})]$

$\delta H^0_ {rxn}=$ ∑$[4*413 kJ)+(614kJ)+(2*647kJ]$+ ∑ $(5*-413kJ)+(-347kJ)+(-467kJ)+(-358kJ)]$

$\delta H^0_ {rxn}=$  -37 kJ

To calculate the heats of reaction by using enthalpies of formation; we have:

$\delta H^0_ {rxn}=$ ∑ $\delta H^0_ {products}$ -  ∑ $\delta H^0_ {reactants}$

$\delta H^0_ {rxn}=$ ∑ $\delta H^0_ {CH_3CH_2OH}$ - ∑ $\delta H^0_ {CH_2=CH_2+H_2O}$

$\delta H^0_ {rxn}=$  (-235.1 kJ) - [(+52.47 kJ) + (-241.826 kJ)]

$\delta H^0_ {rxn}=$  -45.7 kJ